According to my understanding of black hole thermodynamics, if I observe a black hole from a safe distance I should observe black body radiation emanating from it, with a temperature determined by its mass. The energy from this radiation comes from the black hole's mass itself.

But where (in space-time) does the process of generating the Hawking radiation happen? It seems like it should be at the event horizon itself. However, here is a Penrose diagram of a black hole that forms from a collapsing star and then evaporates, which I've cribbed from this blog post by Luboš Motl.

On the diagram I've drawn the world-lines of the star's surface (orange) and an observer who remains at a safe distance and eventually escapes to infinity (green). From the diagram I can see how the observer can see photons from the star itself and any other infalling matter (orange light rays). These will become red-shifted to undetectably low frequencies. But it seems as if any photons emitted from the horizon itself will only be observed at a single instant in time (blue light ray), which looks like it should be observed as the collapse of the black hole.

So it seems that if I observe photons from a black hole at any time before its eventual evaporation, they must have originated from a time before the horizon actually formed. Is this correct? It seems very much at odds with the way the subject of Hawking radiation is usually summarised. How is it possible for the photons to be emitted before the formation of the horizon? Does the energy-time uncertainty relation play a role here?

One reason I'm interested in this is because I'd like to know whether Hawking radiation interacts with the matter that falls in to the black hole. There seem to be three possibilities:

Hawking radiation is generated in the space-time in between the black hole and the observer, and so doesn't interact (much, or at all) with the infalling matter;

Hawking radiation is generated near to the centre of the black hole, at a time before the horizon forms, and consequently it does interact with the matter.

The Hawking radiation is actually emitted by the infalling matter, which for some reason is heated to a very high temperature as it approaches the event horizon.

(With thanks to pjcamp) you can't think of them as coming from a particular point, because they are quantum particles and never have a well-defined location.

All these possibilities have quite different implications for how one should think of the information content of the radiation that eventually reaches the observer, so I'd like to know which (if any) is correct.

The fourth possibility does sound like the most reasonable, but if it's the case I'd like some more details, because what I'm really trying to understand is whether the Hawking photons can interact with the infalling matter or not. Ordinarily, if I observe a photon I expect it to have been emitted by something. If I observe one coming from a black hole, it doesn't seem unreasonable to try and trace its trajectory back in time and work out when and where it came from, and if I do that it will still appear to have come from a time before the horizon formed, and in fact will appear to be originating from surface of the original collapsing star, just before it passed the horizon. I understand the argument that the infalling matter will not experience any Hawking radiation, but I would like to understand whether, from the perspective of the outside observer, the Hawking radiation appears to interact with the matter falling into the black hole. Clearly it does interact with objects that are sufficiently far from the black hole, even if they're free-falling towards it, so if it doesn't interact with the surface of the collapsing star then where is the cutoff point, and why?

In an answer below, Ron Maimon mentions "a microscopic point right where the black hole first formed," but in this diagram it looks like no radiation from that point will be observed until the hole's collapse. Everything I've read about black holes suggests that Hawking radiation is observed to emanate from the black hole continuously, and not just at the moment of collapse, so I'm still very confused about this.

If the radiation is all emitted from this point in space-time, it seems like it should interact very strongly with the in-falling matter:

In this case, crossing the event horizon would not be an uneventful non-experience after all, since it would involve colliding with a large proportion of the Hawking photons all at once. (Is this related to the idea of a "firewall" that I've heard about?)

Finally, I realise it's possible that I'm just thinking about it in the wrong way. I know that the existence of photons is not observer-independent, so I guess it could just be that the question of where the photons originate is not a meaningful one. But even in this case I'd really like to have a clearer physical picture of the situation. If there is a good reason why "where and when do the photons originate?" is not the right question, I'd really appreciate an answer that explains it. (pjcamp's answer to the original version of the question goes some way towards this, but it doesn't address the time-related aspect of the current version, and it also doesn't give any insight as to whether the Hawking radiation interacts with the infalling matter, from the observer's perspective.)

Editorial note: this question has been changed quite a bit since the version that pjcamp and Ron Maimon answered. The old version was based on a time-symmetry argument, which is correct for a Schwartzchild black hole, but not for a transient one that forms from a collapsing star and then evaporates. I think the exposition in terms of Penrose diagrams is much clearer.

@Arjang The event horizon is where the distance from the Black hole equals the Schwarzschild radius.
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DilatonMar 18 '12 at 11:46

@Dilaton : yes, but is it an exact distance or has some level indeterminency? in other words could a photon inside finds itself (if it is sufficeintly close enough to some region we would consider as the event horizon) all of a sudden outside?
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ArjangMar 18 '12 at 11:51

If the photon were emitted outside the black hole, how does the black hole lose mass (energy would have to be transferred across the event horizon and out to the photon). I suspect this implies the photon must be created on the horizon. This is related to my unanswered question here: physics.stackexchange.com/questions/21961/…
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user1247Mar 18 '12 at 12:08

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I thought it was caused by virtual particle pairs coming into existence outside the event horizon, and one of them getting stuck inside while the other radiates.
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endolithMar 19 '12 at 16:21

4 Answers
4

I do not know for your graphic, but as I understand it, you can see Hawking radiation as photons who quantum-tunnel from inside the Black Hole. This is possible because the Hawking radiation wavelength is of the same order as the BH diameter.

As such, you should depict these light-rays as travelling not following the lightcone, but at higher-than light speed through their initial path, and this does not contradict causality because no information can be transmitted from inside the BH (the Hawking photons have strictly thermal spectrum).

Regarding the question of whether this radiation interacts with the infalling observer, the answer is yes, but this interaction is undermined by the fact that this radiation has very large wavelength, exceeding any possible detector the observer could have. So unless the falling onserver has the dimensions comparable with the black hole, he will hardly detect any radiation (yet the probability is not strictly zero).

It's true that by the time they reach a distant observer, the emitted photons have a very long wavelength. But from my point of view as an external observer, if I trace the emitted photons back along their path, I will find that when they were closer to the horizon they had much higher frequency. So (as an outside observer) I should expect them to have been able to interact in the past with much smaller infalling objects than the size of the black hole itself. Ron says it's resolved by the photons having such a high frequency near the horizon that they can't be measured by an observer there.
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NathanielApr 26 '14 at 9:09

@Nathaniel well I encountered in the literature with an opposite claim that the infalling observer cannot detect the photons because he does not have the appropriate detector to measure such high wavelengths.
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AnixxApr 26 '14 at 9:37

The origin of the radiation in Hawking's original calculation is in a microscopic point right where the black hole first formed. This is where your back-tracing procedure ends. Unfortunately, the light is blue shifted insanely during the back-shifting, so that it is far too blue to be physical (it has wavelength past the Planck length by a lot). This made people worry about Hawking's derivation.

The modern picture of holography generally resolves this issue. The black hole is indistinguishable from and quantum mechanically dual to a white hole, so that you can consider the Hawking radiation as coming from the white hole. This is a consistent picture, and the compression of the radiation at the initial point has an exact analog in what happens to infalling matter in a classical white hole, or in an evaporating black hole. Both issues require a duality between the interior and exterior description, and the realization that highly transplanckian objects near the horizon are really best thought of as stringly spread out over the entire surface, or living in a dual interior region.

(The question changed since I answered this, I am getting downvotes: Penrose diagrams are a wrong picture, they are misleading, they are completely incorrect in the proper quantum gravity, don't use it. The proper picture is the naive one, without twisting the horizon to lie at 45 degrees.

When you make the horizon lie at 45 degrees you must choose if the horizon is a past or future horizon. But the two things are quantum-mechanically dual (although classically separate) and you shouldn't force the horizon to be one and not the other. All black holes which are around long enough can be viewed as eternal, and near the end-state, they are all white holes. This was explained by Hawking, and verified by AdS/CFT. Penrose diagrams should be retired, and downvoters, and those who give answers other than mine, have no idea what they're talking about).

The first part of your answer is interesting, thank you. So is it correct to say that in Hawking's original formulation, my option 2 was the correct one: in travelling outward from its instantaneous origin point, the radiation would eventually interact with all the matter that ever fell into the black hole and carry information about it out again? (And hence the information paradox should never have been a serious concern in the first place?)
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NathanielMar 28 '12 at 11:50

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@Nathaniel: the problem is that the "interaction" naively can be made arbitrarily small--- you can look at Hawking radiation of neutrinos. There is no way that the light can be "interacting" with infalling stuff electromagnetically to get the information out. The correct idea is that the interaction is gravitational, and the mechanism is that the horizon has deformation degrees of freedom that encode the infalling information. This is exactly the path by which t'Hooft deduced the holographic principle, and there is no more way to understand this than to understand thermo without entropy.
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Ron MaimonMar 28 '12 at 14:32

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@Nathaniel: This is a start, but neutrinos at high energies actually interact about as strongly as electrons, so it's not a good example. The apparent loss of information during settling to the no-hair state means that you can throw dark-matter in and get photons out. Or throw neutrons in and get electrons/positrons out. This was known well before t'Hooft. The Hawking radiation is paradoxical seeming, because the output is independent of the input semiclassically, and there is no obvious fix by introducing interactions, because the end result is independent of the strength of these.
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Ron MaimonMar 29 '12 at 2:00

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@Dilaton: (and Nathaniel too) The way that the information is imprinted in the horizon is crazy--- its not that the photon is interacting with other stuff frozen on the horizon, its that the horizon itself is jiggling thermally, and the jiggles themselves are the infalling photon and the infalling matter. They contain the entire physics of the infalling matter. This is the holographic idea. When the infalling photon gets close, it is described more and more by BH jiggles (rather than cosmological jiggles), and these jiggles obey an information preserving field theory, at least when extremal.
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Ron MaimonMar 29 '12 at 2:01

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So the process of black hole excitation and decay can be understood only if the black hole itself reconstructs the spacetime near it and around it. This is the near-horizon region in AdS/CFT, where the whole spacetime is reconstructed from the brane theory, with an insane nonlocal mapping that reproduces a gravitational theory in a different number of dimensions. An infalling photon carries a gravitational field, which is determined by the deformation of the black hole horizon ( a boundary condition), and is completely determined when the photon is close to the horizon. This is t'Hooft's stuff
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Ron MaimonMar 29 '12 at 2:04

Nice question! I'll try to put together a few ideas into what might be a valid answer, but this could be wrong.

The event horizon is a lightlike surface. Therefore no proper time passes at the horizon, and any photon emitted exactly at the horizon would remain exactly at the horizon until the moment when evaporation was complete, after which it would fly off -- at the same instant as every other photon that had been trapped there. The Penrose diagram would look like this:

The green world-line represents a distant inertial observer. (Geodesics don't generally look straight on a Penrose diagram.) She receives all of the Hawking radiation (purple) at a single instant. Now we know that this is not really right. Hawking radiation is supposed to be detected at some finite rate by a distant observer, and this rate increases continuously until evaporation is complete.

I think this suggests that we need to consider the concept of a "stretched horizon," which is basically the horizon plus an extra distance on the order of the Planck length. The distant observer applies her knowledge of gravitational time dilation and infers that the horizon is infinitely hot, and therefore that the laws of physics she knows break down there. It thus becomes useless to worry about the exact nature of the degrees of freedom that are present there; they could be at the Planck scale, could be photons, could be strings, could be virtual black holes, could be real black holes that fission off from the main hole by a process analogous to alpha decay. So we just say that there's a sort of "atmosphere" that extends at least one Planck length above the horizon. We then have this picture:

The stretched horizon is the red curve. The purple lines represent photons emitted from it at different times, and detected at different times by the distant observer. I've simply guessed the qualitative shape of the stretched horizon on the Penrose diagram (curved, concave down); maybe someone else could check whether this actually checks out with a specific coordinate transformation.

Thanks, this looks promising. It would seem to imply that from the distant observer's point of view, the photons would appear to come from the practically-infinitely-hot extended horizon, and thus should appear to have interacted very strongly with the infalling matter.
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NathanielMay 14 '13 at 5:59

There's also an interesting issue about the conservation of energy. In your diagram, all of the matter-energy that ever falls in escapes in the form of radiation before the moment of collapse. This seems as if it should be correct. But that seems to imply that none of that matter-energy can ever pass the horizon. I used to think nothing could pass the horizon, but after learning about Penrose diagrams I thought I had been mistaken - now I'm unsure again.
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NathanielMay 14 '13 at 6:04

@Nathaniel: The infinite temperature inferred by a distant observer is not because of any interaction with infalling matter. Hawking radiation isn't even calculated for a black hole that is forming by astrophysical collapse, because that would be too complicated to model. Hawking radiation happens in a purely vacuum (Schwarzschild) spacetime.
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Ben CrowellMay 14 '13 at 15:02

@Nathaniel: In your reasoning about conservation of energy, I think you need to be more careful about how you think about simultaneity. For example, a distant observer and an infalling observer have different notions of time. When you talk about the "moment of collapse," that isn't a time that all observers will agree on.
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Ben CrowellMay 14 '13 at 15:09

I didn't mean to imply that the radiation is caused by interactions with the infalling matter. I'm just saying that if there is any infalling matter then most or all of it be will further away from the event horizon than the region where the Hawking photons appear, and consequently they must interact with it.
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NathanielMay 14 '13 at 15:13

There are a number of equivalent ways to think about Hawking radiation. One is pair creation, as endolith mentions, where the infalling particle has negative total energy and so reduces the mass of the black hole. Another way, perhaps more useful here, involves Compton wavelength. If the wavelength of a particle (not just photons, by the way) is greater than the Schwarzchild radius, then the particle cannot be thought of as localized within the black hole. There is a finite probability that it will be found outside. In other words, you can think of it as a tunneling process. In fact, you can derive the correct Hawking temperature from the correct wavelength and the uncertainty principle, without deploying the full machinery of quantum field theory.

So I guess that counts as #4 because it isn't on your list. You can't think of quantum particles coming from a specific point because you can't think of them ever having a specific location.

Thank you for your answer. I wonder if you could expand it to comment on whether photons emitted in this way interact with the falling matter that an outside observer sees as "frozen" just outside the event horizon?
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NathanielMar 27 '12 at 11:19

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@Nathaniel: The answer to that question is a little more tricky--the Hawking radiation arises because the reference frame of the observer near the horizon is different han that of the observer far from the horizon, and this leads to the two observers having a different notion of what a particle is, which leads to the distant observer saying that the infalling observer's vacuum state actually contains particles. So, the ``frozen'' observer wouldn't actually see any hawking radiation as they fall in--they are freely falling, and see a nice vacuum state, locally.
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Jerry SchirmerMar 27 '12 at 12:45

@JerrySchirmer I can understand that (see this blog post in which I come to a similar conclusion: jellymatter.com/2011/02/26/falling-into-a-black-hole-part-1 - although of course from the infalling observer's point of view there is still a horizon ahead, which should still be radiating) but would you say that, from my point of view as the external observer, the photons would appear to have interacted with the infalling matter? Would they contain any information about it, in other words?
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NathanielMar 27 '12 at 16:06

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@Nathaniel I am not sure if I undestood Nathaniel correctly, but myself I would formulate the problem as follows: Distant observer sees Havking radiation from a black hole (with Planck spectrum and say measurable temperature). Now suppose there is a gas between the event horizon and the observer - would the observer detect absorption line in the black hole spectrum? How does this result change if the gas is free falling? How does it change with distance of the gas (or the observer) from the black hole. Will the distant observer observer the lines if he is free falling?
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Leos OndraMay 16 '13 at 14:17